Courses
Courses for Kids
Free study material
Offline Centres
More
Last updated date: 24th Feb 2024
Total views: 338.1k
Views today: 6.38k
IVSAT 2024
Answer
VerifiedVerified
338.1k+ views
Hint: We take two cases $x+2>0$ and $x+2<0$ . For both the cases, we take the $x+2$ to the right hand side of the inequality and solve for $x$ . We will get one solution in each case, and we will see that for the second case, the solution and the assumption are contradictory.

Complete step by step solution:
The given inequality is
$\dfrac{x-1}{x+2}<0....\left( 1 \right)$
If we closely observe the above inequality, we can see that the function $\dfrac{x-1}{x+2}$ is undefined at $x=-2$ . So, the range of values of $x$ for which the function $\dfrac{x-1}{x+2}$ is defined, or the domain of $x$ for the function $\dfrac{x-1}{x+2}$ is $\mathbb{R}-\left\{ -2 \right\}$ where $\mathbb{R}$ represents the set of all real numbers.
Considering other values of $x$ , we can solve the inequality by taking one time $x+2>0$ and another time $x+2<0$ . Let us first take the case of $x+2>0$ . We now multiply both sides of $\left( 1 \right)$ by $x+2$ and $x+2$ being positive, there is no change in the inequality sign. The inequality thus becomes,
$\Rightarrow x-1<0$
Adding $1$ to both sides of the above inequality, we get,
$\Rightarrow x-1+1<1$
This upon simplifying, we get,
$\Rightarrow x<1$
Before carrying out the operation in this case, we had assumed that $x+2 > 0$ or that $x > -2$ . Thus, the solution for the inequality $\dfrac{x-1}{x+2} < 0$ for the case $x+2 > 0$ is $-2 < x < 1$ .
Let us now consider another case when $x+2<0$ . We now multiply both sides of $\left( 1 \right)$ by $x+2$ and $x+2$ being negative, there will be change in the inequality sign; it changes from $<$ to $>$ , that is, it reverses. The inequality thus becomes,
$\Rightarrow x-1>0$
Adding $1$ to both sides of the above inequality, we get,
$\Rightarrow x-1+1>1$
This upon simplifying, we get,
$\Rightarrow x>1$
Before carrying out the operation in this case, we had assumed that $x+2<0$ or that $x<-2$ . But, upon solving, we are getting $x > 1$ . Both cannot be true at the same time. Thus, there is no solution for this case.
Therefore, we can conclude that the solution for the given inequality is $x\in \left( -2,1 \right)$ .

Note:
While solving problems on inequality, we must consider both the possibilities of positive and negative and solve accordingly. We must not blindly multiply or divide terms on two sides or this will give rise to wrong solutions. We must remember to reverse the sign of the inequality at times of multiplying negative numbers. In the end, we must cross-check our answer and see if it satisfies the conditions or not.
Recently Updated Pages